POJ3349 Snowflake Snow Snowflakes 【哈希表】

题目

很简单，给一堆6元组，可以从任意位置开始往任意方向读，问有没有两个相同的6元组

题解

hash表入门题

先把一个六元组的积 + 和取模作为hash值，然后查表即可

期望\(O(n)\)

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#define LL long long int

#define Redge(u) for (int k = h[u]; k; k = ed[k].nxt)

#define REP(i,n) for (int i = 1; i <= (n); i++)

using namespace std;

const int maxn = 100010,maxm = 100005,INF = 1000000000,P = 100007;

inline int read(){

	int out = 0,flag = 1; char c = getchar();

	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}

	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}

	return out * flag;

}

int h[maxn],nxt[maxn],n,tot;

LL A[6],H[maxn][6];

bool cmp(LL A[],LL B[]){

	for (int pos = 0; pos < 6; pos++){

		bool flag = true;

		for (int i = 0; i < 6; i++)

			if (A[i] != B[(pos + i) % 6]){

				flag = false; break;

			}

		if (flag) return true;

		flag = true;

		for (int i = 0; i < 6; i++)

			if (A[i] != B[(pos - i + 6) % 6]){

				flag = false; break;

			}

		if (flag) return true;

	}

	return false;

}

bool ins(){

	LL s = 0,m = 1,v;

	for (int i = 0; i < 6; i++)

		s = (s + A[i]) % P,m = m * A[i] % P;

	v = (s + m) % P;

	for (int k = h[v]; k != -1; k = nxt[k])

		if (cmp(H[k],A)) return true;

	tot++;

	for (int i = 0; i < 6; i++) H[tot][i] = A[i];

	nxt[tot] = h[v]; h[v] = tot;

	return false;

}

int main(){

	memset(h,-1,sizeof(h));

	n = read();

	for (int i = 1; i <= n; i++){

		for (int j = 0; j < 6; j++)

			A[j] = read();

		if (ins()) {puts("Twin snowflakes found."); return 0;}

	}

	puts("No two snowflakes are alike.");

	return 0;

}

巴特西

POJ3349 Snowflake Snow Snowflakes 【哈希表】

题目

题解

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